I see a mistake on the RHS on the first step:
You have
Check the signs again, I believe it should be:
Find dy/dx for the following:
y^3=x^3/(2-xy)
This is my work.. can someone check to see if I have the correct answer and method? Thanks
d/dx (3y^2)=(3x^2)(2-xy)-x^3[(-1)y+x(1)dy/dx]/(2-xy)^2
d/dx (3y^2)=(3x^2)(2-xy)+x^3y-x(dy/dx)]/(2-xy)^2
(2-xy)^2(3y^2)(dy/dx)+x (dy/dx)=3x^2(2-xy)+x^3y
dy/dx[(2-xy)^2(3y^2)+x]=3x^2(2-xy)+x^3y
dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)+x]
Looks good now ! Just to make sure others understand, make sure to use parenthesis on the bottom:
Yours:
dy/dx=[3x^2(2-xy)+x^3y]/(2-xy)^2(3y^2)-x^4
Yours with parenthesis/brackets:
dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)-x^4]
I'm sure that's what you meant, seeing how your work is correct now.